Question

An electron with initial kinetic energy 5.1 eV encounters a square potential barrier with height 10.0 eV .What is the width of the barrier if the electron has a 0.60 % probability of tunneling through the barrier? Express your answer to two significant figures and include the appropriate units.

Answer 1

`2.9* 10^(-10)\ m`

Step-by-step explanation:

E = Initial kinetic energy = 5.1 eV

`u_0` = Potential barrier height = 10 eV

P = Probability =
`(0.6)/(100)=0.6* 10^(-2)`

m = Mass of electron =
`9.11* 10^(-31)\ kg`

Constant A is given by

`A=16(E)/(u_0)(1-(E)/(u_0))\\\Rightarrow A=16* (5.1)/(10)(1-(5.1)/(10))\\\Rightarrow A=3.9984`

K is given by

`K=(√(2m(u_0-E)))/(\hslash)\\\Rightarrow K=\frac{\sqrt{2* 9.1* 10^(-31)(10-5.1)* 1.6* 10^(-19)}}{1.055* 10^(-34)}\\\Rightarrow K=11322472269.49086\ /m`

We have the relation

`P=Ae^(-2Kl)\\\Rightarrow (P)/(A)=e^(-2Kl)\\\Rightarrow (A)/(P)=e^(2Kl)\\\Rightarrow ln(A)/(P)=2Kl\\\Rightarrow l=(ln(A)/(P))/(2K)\\\Rightarrow l=(ln(3.9984)/(0.6* 10^(-2)))/(2* 11322472269.49086)\\\Rightarrow l=2.87123* 10^(-10)\ m`

The width of the barrier is
`2.9* 10^(-10)\ m`